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Difference between revisions of "2004 AMC 10B Problems/Problem 6"

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Using the fact that <math>n! = n\cdot (n-1)!</math>, we can write:
 
Using the fact that <math>n! = n\cdot (n-1)!</math>, we can write:
 +
 
* <math>A=98! \cdot (99\cdot 98!) = 99 \cdot (98!)^2 = 11\cdot(3\cdot98!)^2</math>
 
* <math>A=98! \cdot (99\cdot 98!) = 99 \cdot (98!)^2 = 11\cdot(3\cdot98!)^2</math>
 +
 
* <math>B=100 \cdot 99 \cdot (98!)^2 = 11\cdot (30\cdot 98!)^2 </math>
 
* <math>B=100 \cdot 99 \cdot (98!)^2 = 11\cdot (30\cdot 98!)^2 </math>
 +
 
* <math>C=100\cdot (99!)^2 = (10\cdot 99!)^2</math>
 
* <math>C=100\cdot (99!)^2 = (10\cdot 99!)^2</math>
 +
 
* <math>D=101\cdot 100\cdot (99!)^2 = 101 \cdot(10\cdot 99!)^2</math>
 
* <math>D=101\cdot 100\cdot (99!)^2 = 101 \cdot(10\cdot 99!)^2</math>
 +
 
* <math>E=101\cdot (100!)^2</math>
 
* <math>E=101\cdot (100!)^2</math>
  

Revision as of 12:46, 20 January 2014

Problem

Which of the following numbers is a perfect square?

$\mathrm{(A) \ } 98! \cdot 99! \qquad \mathrm{(B) \ } 98! \cdot 100! \qquad \mathrm{(C) \ } 99! \cdot 100! \qquad \mathrm{(D) \ } 99! \cdot 101! \qquad \mathrm{(E) \ } 100! \cdot 101!$

Solution

Using the fact that $n! = n\cdot (n-1)!$, we can write:

  • $A=98! \cdot (99\cdot 98!) = 99 \cdot (98!)^2 = 11\cdot(3\cdot98!)^2$
  • $B=100 \cdot 99 \cdot (98!)^2 = 11\cdot (30\cdot 98!)^2$
  • $C=100\cdot (99!)^2 = (10\cdot 99!)^2$
  • $D=101\cdot 100\cdot (99!)^2 = 101 \cdot(10\cdot 99!)^2$
  • $E=101\cdot (100!)^2$

Clearly $\boxed{\mathrm{(C) \ } 99! \cdot 100!}$ is a square, and as $11$, and $101$ are primes, none of the other four are squares.

See also

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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